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Gibbs Free Energy.

  1. Feb 14, 2010 #1
    Hi, ive recently started studying chemistry and have being going over gibbs free energy.. I understand it is a driving force causing chemical and physical changes but during my course of reading I did have some queries stem to mind.

    1) In relation to equilibrium why is it that a reaction occurs such that G is minimum ? I.e. i read that the reason neither the forward nor reverse reaction proceeds to completion during equilibrium is because Gibbs Free Energy is at a minimum. Is this due to no driving force being provided?

    2) When gibbs free energy is positive it indicates the forward reaction (e.g. A --> B) is non-spontaneous why does it still occur? I read something about the reaction still proceeds but the pressure of the product does not read 10^5 P i.e. standard pressure?
    im really confused on this bit can somone please simplify this

    cheers for any help
  2. jcsd
  3. Feb 28, 2010 #2
    Equilibrium can be viewed as compromise between two opposing tendency of the system:
    1. to attain minimum energy (enthalpy)
    2. of maximum molecular chaos (entropy)
    remember this equation,
    where [tex]\Delta[/tex]H is the enthalpy change, [tex]\Delta[/tex]S is the entropy change, and T is the temperature
    A higher value of T[tex]\Delta[/tex]S signifies greater spontaneity. So as this parameter increases, [tex]\Delta[/tex]G becomes smaller and smaller. For Gibbs energy we hav, -[tex]\Delta[/tex]G[tex]\geq[/tex]0. At equilibrium, the two tendencies mentioned above are equal. So, [tex]\Delta[/tex]H=T[tex]\Delta[/tex]S; & [tex]\Delta[/tex]G=0.
    To make a reaction with positive Gibbs energy occur, you just increase the temperature, that will increase the T[tex]\Delta[/tex]S factor in the equation.
    Last edited: Feb 28, 2010
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